If your solution to a stated problem does not match the given solution, you should self-critique per instructions at  

 

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

 

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

 

 

 

004. `query 4

 

 

Question: `q4.6.1 (previously 4.6.06 (was 4.5.06)) y = C e^(kt) thru (3,.5) and (4,5)

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

 

Given Solution:

 

`a Substituting the coordinates of the first and second points into the form y = C e^(k t) we obtain the equations

 

.5 = C e^(3*k)and

5 = Ce^(4k) .

 

Dividing the second equation by the first we get

 

5 / .5 = C e^(4k) / [ C e^(3k) ] or

10 = e^k so

k = 2.3, approx. (i.e., k = ln(10) )

 

Thus .5 = C e^(2.3 * 3)

.5 = C e^(6.9)

C = .5 / e^(6.9) = .0005, approx.

 

The model is thus close to y =.0005 e^(2.3 t). **

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: `q 4.6.2 (previously 4.6.10 (was 4.5.10)) solve dy/dt = 5.2 y if y=18 when t=0

 

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

 

Given Solution:

 

`a The details of the process:

 

dy/dt = 5.2y. Divide both sides by y to get

dy/y = 5.2 dt. This is the same as

(1/y)dy = 5.2dt. Integrate the left side with respect to y and the right with respect to t:

ln | y | = 5.2t +C. Therefore

e^(ln y) = e^(5.2 t + c) so

y = e^(5.2 t + c). This is the general function which satisfies dy/dt = 5.2 y.

 

Now e^(a+b) = e^a * e^b so

y = e^c e^(5.2 t). e^c can be any positive number so we say e^c = A, A > 0.

y = A e^(5.2 t). This is the general function which satisfies dy/dt = 5.2 y.

 

When t=0, y = 18 so

 

18 = A e^0. e^0 is 1 so

A = 18. The function is therefore

 

y = 18 e^(5.2 t). **

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: `q4.6.5 (previously 4.6.25 (was 4.5.25))  Init investment $1000, rate 12%.

 

 

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

 

Given Solution:

 

`a

Rate = .12 and initial amount is $1000 so we have


amt = $1000 e^(.12 t).

The equation for the doubling time is

1000 e^(.105 t) = 2 * 1000.

Dividing both sides by 1000 we get
e^(.12 t) = 2. Taking the natural log of both sides
.12t = ln(2) so that
t = ln(2) / .12 = 5.8 yrs approx.


after 10 years we have

after 25 yrs we have

 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: `q 4.6.8 (previously 4.6.44 (was 4.5.42)) demand fn p = C e^(kx) if when p=$5, x = 300 and when p=$4, x = 400

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

 

Given Solution:

 

`a You get 5 = C e^(300 k) and 4 = C e^(400 k).

 

If you divide the first equation by the second you get

 

5/4 = e^(300 k) / e^(400 k) so

5/4 = e^(-100 k) and

k = ln(5/4) / (-100) = -.0022 approx..

 

Then you can substitute into the first equation:

}

5 = C e^(300 k) so

C = 5 / e^(300 k) = 5 / [ e^(300 ln(5/4) / -100 ) ] = 5 / [ e^(-3 ln(5/4) ] .

 

This is easily evaluated on your calculator. You get C = 9.8, approx.

 

So the function is p = 9.8 e^(-.0022 t). **

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating: